Lemma 61.22.1. Let $k$ be a field. Let $G = \text{Gal}(k^{sep}/k)$ be its absolute Galois group. Further,

1. let $M$ be a profinite abelian group with a continuous $G$-action, or

2. let $\Lambda$ be a Noetherian ring and $I \subset \Lambda$ an ideal an let $M$ be an $I$-adically complete $\Lambda$-module with continuous $G$-action.

Then there is a canonical sheaf $\underline{M}^\wedge$ on $\mathop{\mathrm{Spec}}(k)_{pro\text{-}\acute{e}tale}$ associated to $M$ such that

$H^ i(\mathop{\mathrm{Spec}}(k), \underline{M}^\wedge ) = H^ i_{cont}(G, M)$

as abelian groups or $\Lambda$-modules.

Proof. Proof in case (2). Set $M_ n = M/I^ nM$. Then $M = \mathop{\mathrm{lim}}\nolimits M_ n$ as $M$ is assumed $I$-adically complete. Since the action of $G$ is continuous we get continuous actions of $G$ on $M_ n$. By Étale Cohomology, Theorem 59.56.3 this action corresponds to a (locally constant) sheaf $\underline{M_ n}$ of $\Lambda /I^ n$-modules on $\mathop{\mathrm{Spec}}(k)_{\acute{e}tale}$. Pull back to $\mathop{\mathrm{Spec}}(k)_{pro\text{-}\acute{e}tale}$ by the comparison morphism $\epsilon$ and take the limit

$\underline{M}^\wedge = \mathop{\mathrm{lim}}\nolimits \epsilon ^{-1}\underline{M_ n}$

to get the sheaf promised in the lemma. Exactly the same argument as given in the introduction of this section gives the comparison with Tate's continuous Galois cohomology. $\square$

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